紧急求助：COS根号下xdx的不定积分，要详细步骤

Question

如题所述

Answer 1

令t=√x
x=t^2
dx=2tdt
原式=∫2tcostdt
=2tsint-2∫sintdt
=2tsint+2cost+C
=2√xsin√x+2cos√x+C本回答被提问者采纳

Answer 2

原式=∫cos√xdx
令t=√x
x=t^2
dx=2tdt
原式=∫2tcostdt
=2tsint-2∫sintdt
=2tsint+2cost+C
=2√xsin√x+2cos√x+C

Answer 3

设u=x^(1/2),x=u^2,dx=2udu
∫cosx^(1/2)dx=∫(cosu*2u)du=2∫ud(sinu)
=2u*sinu-2∫sinudu
=2x^(1/2)sin[x^(1/2)]+2cos[x^(1/2)]+C

Answer 4

令t=√x，则x=t²
∴∫cos√Xdx=∫costdt²=2∫tcostdt=2(cost+tsint)+C=2(cos√x+√xsin√x)+C